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JEE Main 20127 MayOnline

Exam: JEE Main
Year: 2012
Date: 7 May
Mode: Online
Subject: Mathematics
Chapter: Circle

JEE Main 2012 · 7 May · Online · Mathematics · Circle

The equation of the circle passing through the point (1,2) and...

The equation of the circle passing through the point

(1,2)

and through the points of intersection of

x^2+y^2-4 x-6 y-21=0

and

3 x+4 y+5=0

is given by

Options

A $x^2+y^2+2 x+2 y+11=0$
B $x^2+y^2-2 x+2 y-7=0$
C $x^2+y^2+2 x-2 y-3=0$
D $x^2+y^2+2 x+2 y-11=0$

Correct answer

D. $x^2+y^2+2 x+2 y-11=0$

Step-by-step solution

Point

(1,2)

lies on the circle

x^2+y^2+2 x+2 y -11=0

, because coordinates of point

(1,2)

satisfy the equation x^2+y^2+2 x+2 y-11=0 Now,

x^2+y^2-4 x-6 y-21=0 \quad

...(i)

\begin{aligned} & x^2+y^2+2 x+2 y-11=0 \\ & 3 x+4 y+5=0 \end{aligned}

From (i) and (iii),

\begin{array}{r} x^2+\left(-\frac{3 x+5}{4}\right)^2-4 x-6\left(-\frac{3 x+5}{4}\right)-21=0 \\ \Rightarrow \quad 16 x^2+9 x^2+30 x+25-64 x \\ +72 x+120-336=0 \end{array}

\begin{aligned} & \Rightarrow \quad 25 x^2+38 x-191=0 \\ & \text{ From (ii) and (iii), } \\ & x^2+\left(-\frac{3 x+5}{4}\right)^2+2 x+2\left(-\frac{3 x+5}{4}\right)-11=0 \\ & \Rightarrow \quad 16 x^2+9 x^2+30 x+25+32 x \end{aligned}

Thus we get the same equation from (ii) and (iii) as we get from equation (i) and (iii). Hence the point of intersections of (ii) and (iii) will be same as the point of intersections of(i) and (iii). Therefore the circle (ii) passing through the point of intersection of circle(i) and point

(1,2)

also as shown in the figure. Hence equation(ii) i.e.

x^2+y^2+2 x+2 y-11=0

is the equation of required circle.

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